So, why does the graph deviate from its proportionality in graph 2 despite the ideal nature of the vapours?
It does not deviate, both curves are the result of the ideal behavior. Lets do some math.
For an ideal binary mixture where: (1) the gas phase behaves as an ideal gas, and (2) the liquid phase behaves as an ideal solution; the LV equilibrium reduces to Raoult's law
$$ y_j P = x_j P_j^\pu{sat} \tag{1} $$
where $P_j^\pu{sat}$ is the saturation pressure of chemical species $j$ at a temperature $T$.
Bubble Curve Summing twice Eq. (1) for both components yields
\begin{align}
&y_1 P + y_2 P = x_1 P_1^\pu{sat} + x_2 P_2^\pu{sat} \\
&(y_1 + y_2) P = x_1 P_1^\pu{sat} + (1 - x_1) P_2^\pu{sat} \\
&\boxed{P(x_1, T) = [P_1^\pu{sat}(T) - P_2^\pu{sat}(T)]x_1 + P_2^\pu{sat}(T)} \tag{2}
\end{align}
Thus, at a fixed temperature, the bubble curve is a linear function of the liquid composition $x_1$. This is the scenario in the left-hand side.
Dew Curve First we put Eq. (1) in a more suitable form to eliminate $x_j$
$$ x_j = \frac{y_jP}{P_j^\pu{sat}} \tag{3} $$
Summing twice Eq. (3) for both components yields
\begin{align}
&x_1 + x_2 = \frac{y_1P}{P_1^\pu{sat}} + \frac{y_2P}{P_2^\pu{sat}} \\
&1 = P\left(\frac{y_1}{P_1^\pu{sat}} + \frac{1 - y_1}{P_2^\pu{sat}}\right) \\
&1 = P\left(\frac{y_1P_2^\pu{sat} + (1 - y_1)P_1^\pu{sat}}
{P_1^\pu{sat}P_2^\pu{sat}}\right) \\
&\boxed{P(y_1,T) = \frac{P_1^\pu{sat}(T)P_2^\pu{sat}(T)}{y_1[P_2^\pu{sat}(T) -
P_1^\pu{sat}(T)] + P_1^\pu{sat}(T)}} \tag{4}
\end{align}
Thus, at a fixed temperature, Eq. (3) has a mathematical equation in the form of $a/(bx + c)$. This is the scenario in the right-hand side.
Therefore in LV equilibrium, as indicated by Raoult's law, the two-phase region is confined between an upper straight line and a lower hyperbolic curve for the $Pxy$ diagram.
